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I am given that the complex function $$f(z)=\frac{(e^{z-1}-1)(\cos(z)-1)}{z^3(z-1)^2}$$ has 2 simple poles, one at $z=0$ and another at $z=1$, and asked to calculate the Residues of the function at the singularities. I know that the residue of a pole $z_0$ of $f(z)$ with order $n$ is given by the formula $\frac{1}{(n-1)!}\lim_{z\rightarrow z_0}(z-z_0)^nf(z)$ and that sometimes using L'Hopital's rule is necessary to calculate the values, however, with the poles in this equation, using L'Hopital's rule seems to make it more difficult.

So far, I've done the following:
$$\text{Res}(f,0)=\lim_{z\rightarrow 0}zf(z)=\lim_{z\rightarrow 0}\frac{(e^{z-1}-1)(\cos(z)-1)}{z^2(z-1)^2}=\lim_{z\rightarrow 0}\frac{(e^{z-1}-1)}{(z-1)^2}\cdot \lim_{z\rightarrow 0}\frac{\cos(z)-1}{z^2}\\=(e^{-1}-1)\lim_{z\rightarrow 0}\frac{\cos(z)-1}{z^2}.$$
From here, I'm not sure how to continue. I checked the answer according to the mark scheme and from this step, the marker jumps to $\text{Res}(f,0)=(e^{-1}-1)\cdot(\frac{-1}{2})$. I can't see where the $\frac{-1}{2}$ has come from.

This happens in a similar way with the residue at $z=1$.
$$\text{Res}(f,1)=\lim_{z\rightarrow 1}(z-1)f(z)=\lim_{z\rightarrow 1}\frac{(e^{z-1}-1)(\cos(z)-1)}{z^3(z-1)}=\lim_{z\rightarrow 1}\frac{e^{z-1}-1}{z-1}\cdot\lim_{z\rightarrow 1}\frac{\cos(z)-1}{z^2}\\=(\cos(1) -1)\cdot\lim_{z\rightarrow 1}\frac{e^{z-1}-1}{z-1}.$$
Again, the marker jumps from this step to $Res(f,1)=(\cos(1)-1))\cdot 1$.

If anyone can help me see how to go from my working out to the answer, that would be much appreciated.

Thank you.

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    $\begingroup$ Your calculations for the residue of $f$ are wrong. You are not applying the formula you presented, and it is not correct either. Another option is using series expansion. $\endgroup$ – Fimpellizieri May 14 '18 at 18:36
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Hint. Once that we note that both poles are of order $1$ (simple poles) then your computations are correct. What you need now is that $$\cos(z)=1-\frac{z^2}{2}+o(z^2)\quad\mbox{and}\quad e^{z-1}=1+(z-1)+o(z-1).$$ Or equivalently, by L'Hopital's rule, $$\lim_{z\rightarrow 0}\frac{\cos(z)-1}{z^2}=\lim_{z\rightarrow 0}\frac{-\sin(z)}{2z}=\lim_{z\rightarrow 0}\frac{-\cos(z)}{2}=-\frac{1}{2} \quad\mbox{and}\quad\lim_{z\rightarrow 1}\frac{e^{z-1}-1}{z-1}=\lim_{z\rightarrow 1}\frac{e^{z-1}}{1}=1.$$

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  • $\begingroup$ I'm not familiar with the $o(z)$ notation. Can you elaborate further about what you mean by it? $\endgroup$ – Fats May 14 '18 at 18:46
  • $\begingroup$ Ah I see what I was doing now. Thank you for your help! $\endgroup$ – Fats May 14 '18 at 18:50
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Hint: If $f$ has a simple pole at $a,$ then

$$\text { Res }(f,a) = \lim_{z\to a}(z-a)f(z).$$

In our case, at $0,$ we have

$$\text { Res }(f,0) = \lim_{z\to 0}zf(z) = \lim_{z\to 0}\frac{(e^{z-1}-1)(\cos z-1)}{z^2(z-1)^2} = \frac{e^{-1}-1}{1^2}\lim_{z\to 0}\frac{\cos z-1}{z^2}.$$

That last limit should look familiar from good old calculus.

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